A zinc-copper battery is constructed as follows at 25°C.
Zn|Zn^2+ (0.15 M)||Cu2+(3.00 M)|Cu
The mass of each electrode is 200. g.
a) Calculate the cell potential when this battery is first connected.
- i got this to be 1.14V
b.) Calculate the cell potential after 10.0 A of current has flowed for 10.0 h
(c) Calculate the mass of each electrode after 10.0 h.
d) How long can this battery deliver a current of 10.0 A before it goes dead?
please help! i just can't seem to figure out how to calculate the concentrations of the Zn^2+ and Cu^2+ after the current has flowed through.
-thanks.
Do you have a volume?
10A x 10hrs x 3600 s/hr = 36,000 coulombs.
96m485 C will dissolve (or plate out) 1 equivalent weight of a metal.
1 equivalent weight of Zn = 65.38/2 = 31.3
1 equivalent weight of Cu = 63.55/2 = 31.8
Zn dissolved = 31.2 x (36,000/96,485) = ??
Cu plated = 31.8 x (36,000/96,485) = ??
?? grams of each can be converted to moles but we need a volume to convert to molarity. I don't see that in the problem.
yes, each half-cell contains 1.00 L of solution so that is the volume
thank you!
To calculate the concentrations of Zn^2+ and Cu^2+ after the current has flowed through, we can use the concept of Faraday's Law of Electrolysis.
To begin with, let's understand the initial setup of the battery. In the given zinc-copper battery, the Zn electrode is connected to the Zn^2+ ion and the Cu^2+ ion is connected to the Cu electrode. The two half-cells are separated by a salt bridge, represented as "||".
(a) To calculate the cell potential when the battery is first connected, we need to use the Nernst equation for each half-cell:
Ecell = E°cell - (RT/nF) ln(Q)
E°cell is the standard cell potential, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25°C + 273.15 = 298.15 K), n is the number of electrons transferred per mole of reaction, F is Faraday's constant (96,485 C/mol), and Q is the reaction quotient.
For the zinc half-cell:
Zn → Zn^2+ + 2e^-
E°Zn = -0.76 V (given)
QZn = [Zn^2+]/[Zn]^2 (concentration ratio)
For the copper half-cell:
Cu^2+ + 2e^- → Cu
E°Cu = 0.34 V (given)
QCu = [Cu]/[Cu^2+]^2 (concentration ratio)
Plug in the values into the Nernst equation for each half-cell, and sum them up:
Ecell = E°Zn - (RT/2F) ln([Zn^2+]/[Zn]^2) + E°Cu - (RT/2F) ln([Cu]/[Cu^2+]^2)
Substituting the values, we have:
Ecell = -0.76 V - (8.314 J/(mol·K) * 298.15 K / (2 * 96,485 C/mol)) ln([Zn^2+]/[Zn]^2) + 0.34 V - (8.314 J/(mol·K) * 298.15 K / (2 * 96,485 C/mol)) ln([Cu]/[Cu^2+]^2)
Calculating this expression will give you the cell potential when the battery is first connected. In this case, you've already found it to be 1.14 V.
(b) To calculate the cell potential after 10.0 A of current has flowed for 10.0 h, we'll use Faraday's Law of Electrolysis. The total charge passed through the circuit can be calculated using the equation:
Q = It
Where Q is the charge in Coulombs, I is the current in Amperes, and t is the time in seconds.
In this case, I = 10.0 A and t = 10.0 h = 10.0 * 60 * 60 s = 36,000 s.
Now, divide the total charge (Q) by Faraday's constant (F) to find the total number of moles of electrons transferred.
n = Q / F
Next, determine the moles of Zn^2+ and Cu^2+ ions produced or consumed during electrolysis using the stoichiometry of the balanced redox reactions:
Zn → Zn^2+ + 2e^-
1 mol Zn -> 1 mol Zn^2+ + 2 mol e^-
Cu^2+ + 2e^- → Cu
1 mol Cu^2+ + 2 mol e^- -> 1 mol Cu
Now, multiply the number of moles of electrons by the stoichiometric coefficients to determine the moles of the respective ions produced or consumed in the half-cells.
Finally, calculate the new concentrations of Zn^2+ and Cu^2+ ions by dividing the number of moles by the appropriate volume.
(c) To calculate the mass of each electrode after 10.0 h, we need to determine the change in mass caused by the deposition or dissolution of Zn and Cu during electrolysis.
We can calculate the moles of Zn and Cu using the same method mentioned in part (b). Then, we need to multiply the moles of Zn and Cu by their respective molar masses to determine the change in mass.
(d) To determine how long the battery can deliver a current of 10.0 A before it goes dead, we can use the concept of Faraday's Law of Electrolysis again. We need to calculate the total amount of charge needed to deplete all the Zn and Cu ions.
Once we have the total charge required, we can use the same equation Q = It to determine the time required to deliver that amount of charge, with the current I equal to 10.0 A.
10A x 10hrs x 3600 s/hr = 36,000 coulombs.
You forgot to multiply by 10A...it should be 360,000 coulombs. 36,000 is seconds in 10 hours.